Two point charges $+8q$ and $-2q$ are located at $x=0$ and $x=L$ respectively. The location of a point on the x-axis at which the net electric field due to these two point charges is zero is:

For the net field to be zero, the point must be outside the two charges, closer to the smaller charge ($-2q$).Let the point be at a distance $r$ from $-2q$ (so it is at $L+r$ from $+8q$).
$$E_{8q}=E_{2q}$$
$$\frac{k(8q)}{(L+r{)}^2}=\frac{k(2q)}{r^2}$$
$$\frac{4}{(L+r{)}^2}=\frac{1}{r^2}$$
Taking square root:
$$\frac{2}{L+r}=\frac{1}{r}⟹2r=L+r⟹r=L$$
The location from the origin ($x=0$) is $x=L+r=L+L=2L$.